# Ellipse

## Definition and Properties of an Ellipse

An ellipse is a closed plane curve such that the sum of the distances from any point of the curve to two other fixed points (called the foci of the ellipse) is constant. The midpoint of the line segment joining the foci is called the center of the ellipse. The distance from the center to a focus is called the focal distance and denoted by c.

Any ellipse has two axes of symmetry: the first or focal line passing through the foci and the second line perpendicular to the first axis. The points of intersection of these axes with the ellipse are called the vertices. A line segment that runs from the center of the ellipse to its vertex is called the semi-axis of the ellipse. The semi-major axis is usually denoted by a, and the semi-minor axis is denoted by b.

An ellipse with the center at the origin and the semi-axes lying on the coordinate lines is described by the following canonical or standard equation:

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

The sum of the distances from any point of an ellipse to its foci is constant:

$r_1 + r_2 = 2a$

where $$r_1,$$ $$r_2$$ are the distances from an arbitrary point $$P\left({x,y}\right)$$ of the ellipse to the foci $$F_1$$ and $$F_2,$$ $$a$$ is the semi-major axis of the ellipse.

Relationship between the semi-axes of an ellipse and its focal distance:

$a^2 = b^2 + c^2$

where $$a$$ is the semi-major axis of the ellipse, $$b$$ is the semi-minor axis, $$c$$ is the focal distance.

## Eccentricity of an Ellipse

$e = \frac{c}{a} \lt 1$

## Equation of the Directrix of an Ellipse

The directrix of an ellipse is a straight line perpendicular to the focal axis of the ellipse and intersecting it at the distance $$\frac{a}{e}$$ from the center. An ellipse has two directrices spaced on opposite sides of the center. The equations of the directrices are written in the form

$x = \pm\frac{a}{e} = \pm\frac{a^2}{c}$

## Equation of an Ellipse in Parametric Form

\left\{ \begin{aligned} x &= a\cos t \\ y &= b\sin t \end{aligned} \right.,\;\; 0 \le t \le 2\pi

where $$a$$, $$b$$ are the semi-axes of the ellipse, $$t$$ is a parameter.

## General Equation of an Ellipse

$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$

where $$A,$$ $$B,$$ $$C,$$ $$D,$$ $$E,$$ $$F$$ are real numbers and $${B^2 - 4AC \lt 0}.$$

## General Equation of an Ellipse with Semi-Axes Parallel to the Coordinate Axes

$Ax^2 + Cy^2 + Dx + Ey + F = 0$

where $${AC \gt 0}.$$

## Circumference of an Ellipse

$L = 4aE\left(e\right)$

where $$a$$ is the semi-major axis, $$e$$ is the eccentricity, $$E$$ is the complete elliptic integral of the second kind.

## Approximate Formulas for the Circumference of an Ellipse

$L \approx \pi \left[{\frac{3}{2}\left({a+b}\right) - \sqrt{ab}}\right],\;L \approx \pi\sqrt{2\left({a^2 + b^2}\right)}$

where $$a, b$$ are the semi-axes of the ellipse.

## Area of an Ellipse

$S = \pi ab$

## Solved Problems

### Example 1.

Find the semi-axes, foci and eccentricity of the ellipse $${9x^2 + 25y^2 = 225.}$$

Solution.

First we re-write the equation of the ellipse in standard form:

$9x^2 + 25y^2 = 225, \Rightarrow \frac{9x^2}{225} + \frac{25y^2}{225} = 1, \Rightarrow \frac{x^2}{\frac{225}{9}} + \frac{y^2}{\frac{225}{25}} = 1,$
$\Rightarrow \frac{x^2}{25} + \frac{y^2}{9} = 1, \Rightarrow \frac{x^2}{5^2} + \frac{y^2}{3^2} = 1.$

We see that the length of the semi-major axis is $$a = 5$$, and the length of the semi-minor axis is $$b = 3.$$

The center of the ellipse is at the origin. The distance from the center to a focus is determined by the formula

$c^2 = a^2 - b^2.$

So

$c = \sqrt{a^2 - b^2} = \sqrt{5^2 - 3^2} = \sqrt{16} = 4,$

that is, the coordinates of the foci are equal

$F_1 = \left({-c,0}\right) = \left({-4,0}\right);\;\;F_2 = \left({c,0}\right) = \left({4,0}\right).$

Let's calculate the eccentricity of the ellipse:

$e = \frac{c}{a} = \frac{4}{5}.$

### Example 2.

Write the equation of an ellipse if it is known that the distance between the foci is $$16$$ and the semi-minor axis is $$b = 6.$$

Solution.

According to the condition, the distance between the foci is $$16,$$ that is

$2c = 16, \Rightarrow c = 8.$

Find the semi-major axis of the ellipse:

$c^2 = a^2 - b^2, \Rightarrow a^2 = c^2 + b^2, \Rightarrow a = \sqrt{c^2 + b^2} = \sqrt{8^2 + 6^2} = \sqrt{100} = 10.$

Then the standard equation of the ellipse equation is written as

$\frac{x^2}{10^2} + \frac{y^2}{6^2} = 1, \Rightarrow \frac{x^2}{100} + \frac{y^2}{36} = 1.$

### Example 3.

Write the equation of an ellipse passing through the points $$M_1\left({1,2}\right)$$ and $$M_2\left({2,0}\right).$$

Solution.

We will assume that the center of the ellipse is at the origin. Such an ellipse is described by the standard equation

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.$

Let's substitute the coordinates of the given points and get a system of two equations:

$\left\{ \begin{array}{l} \frac{1^2}{a^2} + \frac{2^2}{b^2} = 1 \\ \frac{2^2}{a^2} + \frac{0^2}{b^2} = 1 \end{array} \right., \Rightarrow \left\{ \begin{array}{l} \frac{1}{a^2} + \frac{4}{b^2} = 1 \\ \frac{4}{a^2} + 0 = 1 \end{array} \right..$

Let's find $$a^2$$ from the second equation:

$\frac{4}{a^2} = 1, \Rightarrow a^2 = 4.$

Substituting into the first equation, we find $$b^2:$$

$\frac{1}{4} + \frac{4}{b^2} = 1, \Rightarrow \frac{4}{b^2} = 1 - \frac{1}{4} = \frac{3}{4}, \Rightarrow 3b^2 = 16, \Rightarrow b^2 = \frac{16}{3}.$

Now we can write the equation of the ellipse:

$\frac{x^2}{4} + \frac{y^2}{\frac{16}{3}} = 1,\;\;\text{or}\;\;\frac{x^2}{4} + \frac{3y^2}{16} = 1.$

### Example 4.

Write the equations of the directrices of the ellipse $$x^2 + 4y^2 = 1.$$

Solution.

Let us represent the equation of the ellipse in the form

$x^2 + 4y^2 = 1, \Rightarrow \frac{x^2}{1} + \frac{y^2}{\frac{1}{4}} = 1, \;\text{or}\;\frac{x^2}{1^2} + \frac{y^2}{\left({\frac{1}{2}}\right)^2} = 1,$

that is, the semi-axes of the ellipse are $$a = 1,$$ $$b = \frac{1}{2}.$$

Remember that the equations of the directrices are written as

$x = \pm \frac{a^2}{c} = \pm \frac{a^2}{\sqrt{a^2 - b^2}}.$

Substituting the semi-axes $$a$$ and $$b,$$ we find the equations of the directrices:

$x = \pm \frac{1^2}{\sqrt{1^2 - \left({\frac{1}{2}}\right)^2}} = \pm \frac{1}{\sqrt{1 - \frac{1}{4}}} = \pm\frac{1}{\sqrt{\frac{3}{4}}} = \pm \frac{2}{\sqrt{3}}.$

### Example 5.

In an ellipse, the distance between the foci is equal to the arithmetic mean of the lengths of its axes. Find the eccentricity of the ellipse.

Solution.

The distance between the foci is $$2c.$$ By condition, we know that

$2c = \frac{2a + 2b}{2} = a + b, \Rightarrow c = \frac{a + b}{2}.$

Since $$c^2 = a^2 - b^2,$$ we have

$\left({\frac{a + b}{2}}\right)^2 = a^2 - b^2, \Rightarrow \frac{\left({a + b}\right)^{\cancel{2}}}{4} = \left({a - b}\right)\cancel{\left({a + b}\right)}, \Rightarrow = a + b = 4\left({a - b}\right).$

In the last equation, divide both sides by $$a$$ and determine the ratio of the semi-axes:

$1 + \frac{b}{a} = 4\left({1 - \frac{b}{a}}\right), \Rightarrow 1 + \frac{b}{a} = 4 - \frac{4b}{a}, \Rightarrow \frac{5b}{a} = 3, \Rightarrow \frac{b}{a} = \frac{3}{5}.$

Now let's calculate the eccentricity of the ellipse:

$e = \frac{c}{a} = \frac{a + b}{2a} = \frac{1}{2}\left({1 + \frac{b}{a}}\right) = \frac{1}{2}\left({1 + \frac{3}{5}}\right) = \frac{1}{2} \cdot \frac{8}{5} = \frac{4}{5}.$

### Example 6.

A meridian of the globe has the shape of an ellipse whose axes are in a $$\frac{599}{600}$$ ratio. Determine the eccentricity of the earth's meridian.

Solution.

The eccentricity of an ellipse is expressed by the formula

$e = \frac{c}{a},$

where the distance from the center to a focus is given by

$c^2 = a^2 - b^2.$

Substituting this into the above formula, we get

$e = \frac{\sqrt{a^2 - b^2}}{a} = \sqrt{\frac{a^2-b^2}{a^2}} = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \left({\frac{b}{a}}\right)^2}.$

Let's substitute the ratio of the semi-axes $$\frac{b}{a} = \frac{599}{600}$$ and calculate the eccentricity of the Earth:

$e = \sqrt{1 - \left({\frac{599}{600}}\right)^2} = \sqrt{1 - \frac{358801}{360000}} = \sqrt{\frac{1199}{360000}} \approx \sqrt{0,0033} \approx 0,058$